What is the derivative of (4x^3+9/x) ?

The derivative of (4x^3+9/x) is (8x^3-9)/(x^2)

What is the first derivative of (4x^3+9/x) ?

The first derivative of (4x^3+9/x) is (8x^3-9)/(x^2)

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\frac{d}{d x} (\frac{4 x ^{3} + 9}{x})

\frac{8 x ^{3} - 9}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 x ^{3} + 9}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( 4 x ^{3} + 9 ) x - \frac{d x}{d x} ( 4 x ^{3} + 9 )}{x ^{2}}

\frac{d}{d x} (4 x^{3} + 9) = 12 x^{2}

\frac{d x}{d x} = 1

= \frac{12 x ^{2} x - 1 \cdot ( 4 x ^{3} + 9 )}{x ^{2}}

Simplify

\frac{12 x ^{2} x - 1 \cdot ( 4 x ^{3} + 9 )}{x ^{2}} : \frac{8 x ^{3} - 9}{x ^{2}}

= \frac{8 x ^{3} - 9}{x ^{2}}

\frac{d}{d t} (sin (ω t + θ))

\int_{0}^{1} (1) d x

x e^{x} y^{^{'}} + 2 e^{x} y = 0

\int 9 - 16 x^{2} d x

What is the derivative of (4x^3+9/x) ?
The derivative of (4x^3+9/x) is (8x^3-9)/(x^2)
What is the first derivative of (4x^3+9/x) ?
The first derivative of (4x^3+9/x) is (8x^3-9)/(x^2)